Integrand size = 22, antiderivative size = 56 \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\frac {\arctan \left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}}-\frac {\text {arctanh}\left (\sqrt [4]{2} \sqrt [4]{1+2 x+2 x^2}\right )}{2^{3/4}} \] Output:
1/2*arctan(2^(1/4)*(2*x^2+2*x+1)^(1/4))*2^(1/4)-1/2*arctanh(2^(1/4)*(2*x^2 +2*x+1)^(1/4))*2^(1/4)
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\frac {\arctan \left (\sqrt [4]{2+4 x+4 x^2}\right )-\text {arctanh}\left (\sqrt [4]{2+4 x+4 x^2}\right )}{2^{3/4}} \] Input:
Integrate[1/((1 + 2*x)*(1 + 2*x + 2*x^2)^(1/4)),x]
Output:
(ArcTan[(2 + 4*x + 4*x^2)^(1/4)] - ArcTanh[(2 + 4*x + 4*x^2)^(1/4)])/2^(3/ 4)
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.62, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1118, 27, 243, 73, 25, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(2 x+1) \sqrt [4]{2 x^2+2 x+1}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt [4]{2}}{(2 x+1) \sqrt [4]{(2 x+1)^2+1}}d(2 x+1)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{(2 x+1) \sqrt [4]{(2 x+1)^2+1}}d(2 x+1)}{2^{3/4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \frac {1}{(2 x+1)^2 \sqrt [4]{2 x+2}}d(2 x+1)^2}{2\ 2^{3/4}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \sqrt [4]{2} \int -\frac {(2 x+1)^4}{1-(2 x+1)^8}d\sqrt [4]{2 x+2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\sqrt [4]{2} \int \frac {(2 x+1)^4}{1-(2 x+1)^8}d\sqrt [4]{2 x+2}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \sqrt [4]{2} \left (\frac {1}{2} \int \frac {1}{(2 x+1)^4+1}d\sqrt [4]{2 x+2}-\frac {1}{2} \int \frac {1}{1-(2 x+1)^4}d\sqrt [4]{2 x+2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \sqrt [4]{2} \left (\frac {1}{2} \arctan \left (\sqrt [4]{2 x+2}\right )-\frac {1}{2} \int \frac {1}{1-(2 x+1)^4}d\sqrt [4]{2 x+2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \sqrt [4]{2} \left (\frac {1}{2} \arctan \left (\sqrt [4]{2 x+2}\right )-\frac {1}{2} \text {arctanh}\left (\sqrt [4]{2 x+2}\right )\right )\) |
Input:
Int[1/((1 + 2*x)*(1 + 2*x + 2*x^2)^(1/4)),x]
Output:
2^(1/4)*(ArcTan[(2 + 2*x)^(1/4)]/2 - ArcTanh[(2 + 2*x)^(1/4)]/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
Time = 1.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {1}{4}} \left (2 \arctan \left (2^{\frac {1}{4}} \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}\right )-\ln \left (\frac {2^{\frac {3}{4}}+2 \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}}{-2^{\frac {3}{4}}+2 \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}}\right )\right )}{4}\) | \(70\) |
Input:
int(1/(2*x+1)/(2*x^2+2*x+1)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/4*2^(1/4)*(2*arctan(2^(1/4)*(2*x^2+2*x+1)^(1/4))-ln((2^(3/4)+2*(2*x^2+2* x+1)^(1/4))/(-2^(3/4)+2*(2*x^2+2*x+1)^(1/4))))
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\frac {1}{8} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{4} \cdot 8^{\frac {3}{4}} {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{16} \cdot 8^{\frac {3}{4}} \log \left (8^{\frac {1}{4}} + 2 \, {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} \cdot 8^{\frac {3}{4}} \log \left (-8^{\frac {1}{4}} + 2 \, {\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}}\right ) \] Input:
integrate(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x, algorithm="fricas")
Output:
1/8*8^(3/4)*arctan(1/4*8^(3/4)*(2*x^2 + 2*x + 1)^(1/4)) - 1/16*8^(3/4)*log (8^(1/4) + 2*(2*x^2 + 2*x + 1)^(1/4)) + 1/16*8^(3/4)*log(-8^(1/4) + 2*(2*x ^2 + 2*x + 1)^(1/4))
\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int \frac {1}{\left (2 x + 1\right ) \sqrt [4]{2 x^{2} + 2 x + 1}}\, dx \] Input:
integrate(1/(1+2*x)/(2*x**2+2*x+1)**(1/4),x)
Output:
Integral(1/((2*x + 1)*(2*x**2 + 2*x + 1)**(1/4)), x)
\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}} {\left (2 \, x + 1\right )}} \,d x } \] Input:
integrate(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((2*x^2 + 2*x + 1)^(1/4)*(2*x + 1)), x)
\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{4}} {\left (2 \, x + 1\right )}} \,d x } \] Input:
integrate(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x, algorithm="giac")
Output:
integrate(1/((2*x^2 + 2*x + 1)^(1/4)*(2*x + 1)), x)
Timed out. \[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int \frac {1}{\left (2\,x+1\right )\,{\left (2\,x^2+2\,x+1\right )}^{1/4}} \,d x \] Input:
int(1/((2*x + 1)*(2*x + 2*x^2 + 1)^(1/4)),x)
Output:
int(1/((2*x + 1)*(2*x + 2*x^2 + 1)^(1/4)), x)
\[ \int \frac {1}{(1+2 x) \sqrt [4]{1+2 x+2 x^2}} \, dx=\int \frac {1}{2 \left (2 x^{2}+2 x +1\right )^{\frac {1}{4}} x +\left (2 x^{2}+2 x +1\right )^{\frac {1}{4}}}d x \] Input:
int(1/(1+2*x)/(2*x^2+2*x+1)^(1/4),x)
Output:
int(1/(2*(2*x**2 + 2*x + 1)**(1/4)*x + (2*x**2 + 2*x + 1)**(1/4)),x)